digital communications cheat sheet

Nasser M. Abbasi

July 2, 2015 page compiled on July 2, 2015 at 5:57pm

1 What is the relation between bandpass, baseband ,complex envelop and pre envelop?


2 Some useful Fourier Transforms

x(t)  X (f)
sin (2 πf t)
        c  1-[δ(f − f ) − δ(f + f )]
2j        c           c
cos (2πfct)  1
2 [δ(f − fc) + δ(f + fc)]
cos (2πfct + 𝜃)   [                            ]
12 ej𝜃δ(f −  fc) + e−j𝜃δ(f + fc)
sin (2 πfct + 𝜃 )   [                            ]
12 ej𝜃δ(f −  fc) − e−j𝜃δ(f + fc)

3 Random process definitions

X  (t)  is Stationary: If all its statistics do not change with shift of origin

X (t)  is Wide Sense Stationary: If the mean is constant, and R  (t,t + τ) = R (τ )
  x             x

where autocorelation Rx (τ)  is defined as         ∗
E [x (t)x  (t + τ)].  Note, if X (t)  is real, then Rx (τ)  is real and even

Note: R (x)  must be WSS if it is ergodic.  So ergodic process has constant mean.

4 How to determine Hilbert transform of a signal?

input x (t)  . Find ˆx(t)  which is Hilbert transform of x (t)  defined as                1
ˆx (t) = x (t) ⊗ πt

An easy way is to first find ˆG (f)  which is the Fourier transform of ˆx (t)  and then inverse it to find ˆx (t)

ˆG (f) = − j sgn (f ) G (f)

Where G (f)  is Fourier transform of x (t)

5 How to find Power Spectrum (PSD) of a random signal x (t)

input: random signal x (t)

output: PSD of x (t)


Find autocorrelation Rx (τ)  of x(t)
Find the Fourier Transform of R  (τ)
  x  . The result is the PSD of x (t)  called S  (f)

Another method (this below works if not random x (t)  ) , why? can;t find FT for random process?

Find Fourier Transform X (f)  of x (t)
Find the        2           ∗
|X (f )| = X (f )X  (f)

6 What is the relation between variance and power for a random signal x (t)  ?

Variance is the sum of the total average normalized power and the DC power.

      total Power  DC power
      ◜-[◞◟---◝]   ◜--◞◟--◝2
σ2x = E  x2(t) +  E [x (t)]

For the a signal whose mean is zero,

     ◜tot[al◞ P◟ower◝]
σ2 = E  x2 (t)

How to find average, power, PEP, effective value (or the RMS) of a periodic function?

Let x(t)  be a periodic function, of period T  , then

                            1 ∫ T
average of x (t) = ⟨x (t)⟩ = --     x(t)dt
                           T   0

The average power is

                  ∫  T
pav = ⟨x2(t)⟩ = 1-    |x(t)|2dt
                T   0

Effective value, or the RMS value is

                              ∘  --------------
          ∘  -------  √ ---      1 ∫ T
xrms (t) =    ⟨x2 (t)⟩ =  pav =    --    x2 (t)dt
                                 T  0

For example, for                                   A2
x(t) = A cos(x ),⟨x(t)⟩ = 0,Pav = -2-,xrms(t) = 0.707A

To find PEP (which is the peak envelope power), find the complex envelope ˜x (t)  , then find the average power of it. i.e.

PEP   = 1-˜x2  (t)
        2  max

7 How to find the SNR for sampling quantization?

Suppose we have a message m (t)  that is sampled. Assume we have n  bits to use for encoding the sample levels. Hence there are 2n  levels of quantizations. We want to find the ration of the signal to the noise power. Noise here is generated due to quantization (i.e. due to the rounding off values of m  (t)  during sampling).

This is the algorithm:

Input: n  , the number of bits for encoding, mp  absolute maximum value of the message m (t)  , the pdf fX (t)  of the message m (t)  is m  (t)  is random message or m (t)  function if it is deterministic (such as cos(t) )

Find the quantization step size      2mp--
S =  22
Find Pav  of the error is -1  2
12S   where S  is the step size found in (1), hence                 (    )
P   = -1S2 =  1-  2mp--2 = --m2p-
 av   12      12  22      3×22n
If m (t)  is deterministic find          2       1-∫T       2
pav = ⟨m  (t)⟩ = T  0 |m (t)|dt
If m  (t)  is random, find                   ∫
pav = E (m  (t)) =   m2  (t) fX (t) dt  , this is called the second moment of the pdf
        E (m (t))
SN R  = --m2p--

Hence find SN R  for noise quantisation comes down to finding the power in the message m (t)  .

Examples: For sinosoidal message m (t)  , SN Rdb =  6n + 1.761  . For random m (t)  with PDF which is uniform distributed SN  Rdb = 6n  , for random m (t)  which is AWGN. Do this later

8 How to determine coding of a number from quantization?

Given an analog value say x  and given a maximum absolute possible value to be mp  , and given the number of bits available for coding to be N  , the following are the algorithm to generate the quantiazed version of      x  , called ˆx

8.1 sign mangnitude

Input: x, mp, N

output: ˆx

Let Δ =  2mNp−1   called the step size

Let           [     ]
q = round  abs(x)
             Δ which is the quantization level

If q ≥ 2N −1 − 1  then q = 2N −1   end if  

if x < 0  then code =  q + 2N −1   else code = q  endif

return code  in base 2

8.2 ones complement

Input: x, mp, N

output: ˆx

Let Δ =  -mNp−1
     2   called the step size

Let           [abs(x)]
q = round    Δ which is the quantization level

If q ≥ 2N −1 − 1  then q = 2N −1 − 1  end if

If x > 0  then code = q  else         ( N    )
code =   2  − 1  − q  endif

return code  in base 2

8.3 offset binary

Input: x, mp, N

output: ˆx

Let      -mp-
Δ =  2N−1   called the step size

Let           [     ]
q = round  abs(x)
             Δ which is the quantization level

If       Δ
x ≥ − 2-   then

  if q ≥ 2N −1 − 1  then

q = 2N −1 − 1

  end if

          N −1
code = 2    +  q


  if q ≥ 2N −1 − 1  then

q = 2N −1

  end if

          N −1
code = 2    −  q

end if

return code  in base 2

8.4 2's complement

Input: x, mp, N

output: ˆx

Let       m
Δ =  2Np−1   called the step size

Let           [     ]
q = round    Δ which is the quantization level

If x ≥ − Δ2-   then

  if q ≥ 2N −1 − 1  then

q = 2N −1 − 1

  end if

  code = q


  if q ≥ 2N −1 − 1  then

     N −1
q = 2

  end if

  code = 2N −  q

end if

return code  in base 2

9 How to derive the Phase and Frequency modulation signals?

For any bandpass signal, we can write it as

          (         )
x(t) = Re  ˜x (t) ejωct

Where ˜x (t)  is the complex envelope of x(t)  . For PM and FM, the baseband modulated signal, ˜x (t)  has the form     j𝜃(t)
Ace    Hence the above becomes


But cos (A + B ) = cosA cos B − sinA sin B  , hence the above becomes

x (t) = cos (ωct + 𝜃 (t))

The above is the general form for PM and FM. Now, for PM, 𝜃(t) = k m (t)
        p  and for FM, 𝜃 (t) = k  ∫tm  (t) dt
        f  0    1    1   . Hence, substituting in (1) we obtain

              (         ∫ t         )

xF M (t) = cos  ωct + kf 0 m  (t1) dt1


x    (t) = cos (ω t + k m (t))
 P M            c     p

10 How to obtain the phase deviation and the frequency deviation for angle modulated signal?

From the general form for angle modulated signal (see above note)

x (t) = cos (ωct + 𝜃 (t))

The phase deviation is 𝜃(t)  . And the maximum phase deviation is simply the maximum of 𝜃 (t)

Now, to find the frequency deviation, we need a little bit more work. Start with

fi = fc + Δf

Where fi  is the instantaneous frequency in Hz. But


11 How to quickly determine SNRi  from SN  R
     c  ?

First find SN  Rc  , for to find SN Ri  use the following

SN Ri =  SN Rc BT-   , where BT  is the transmission bandwidth, and B  is the baseband bandwidth. For AM  , BT  = 2B  . For DSB   − SC  , BT  = 2B  . For DSB   − SS  , BT =  B.


12 How to determine figure of merit for DSB-SC using coherent detector?

Figure of merit, γ  is defined as SNRo-
SNRc   where SN  Ro  is the signal-to-noise ratio on output from modulator, and SN Rc  is signal-to-noise ratio for the channel, assuming channel has AWGN added. The following diagram shows the calculations. I used a coherent demodulator.


Question: Verify the above.

13 How to determine figure of merit for AM transmission using coherent detector?


14 How to determine figure of merit for AM using envelope detector?


s1 (t) = Ac (1 + kam (t))cosωct

s2(t) = Ac (1 + kam (t))cos ωct + w (t)


Now assuming ⟨m (t)⟩ = 0  , the above simplifies to

         Ac2-(1 + k2aPm )
SN Rc =  --------------



Now find s3 (t)


Now, to find s4(t)  , which is the envelope of s3 (t) .


Now, assuming Ac ≫  |nI (t)| and Ac ≫  |nQ (t)| , then the above simplifies to

s4(t) = Ac (1 + kam (t)) + nI (t)

now apply the DC blocker, we obtain

s5(t) = Ackam (t) + nI (t)

         ⟨            ⟩
          (Ackam  (t))2     A2ck2aPm
SN Ro =  -------2-------=  --------
            E [nI (t)]      2BN0

                A2ck2aPm-        2
γ = SN--Ro-=  ---2BN0----=  -k-aPm---
    SN  Rc    A2c2 (1+k2aPm)   1 + k2aPm

We notice, that for Large SN Ri  , this detector gives the same result as coherent detector.

For small SN  Ri  , it is better to use the coherent detector than the envelope detector.

15 How to determine figure of merit for SSB using coherent detector?


The difference here is that SSB signal has transmission bandwidth BT  = B  and not 2B  as in all the previous signals. Assume we are working with upper sideband. Analysis is the same for lower sideband.

s1(t) = k[m (t)cos ωct − ˆm (t)sin ωct]

Where k  is a constant. Usually Ac
2   but we will leave it as k  for now. mˆ (t)  is the Hilbert transform of m  (t)

s2 (t) = k [m  (t) cosωct − ˆm (t)sinωct] + w (t)


Assume ⟨m  (t)⟩ = 0  , we obtain

s3 (t) = k [m (t)cosωct − ˆm (t)sinωct] + n (t)



After low pass filter, we obtain

        1  ′         A ′
s5(t) = -A ckm (t) + --cnI (t)
        2             2







16 How to determine figure of merit for VSB using coherent detector?

s(t) = Ac-[m (t)cosω  t ∓ m  (t)sin ω t]
        2            c     Q        c

mQ (t)  is the output of VSB filter when input is m (t)